Question

The function f(x)=125(0.9)x models the population of a species of fly in millions after x years.

How does the average rate of change between years 11 and 15 compare to the average rate of change between years 1 and 5?

The average rate of change between years 11 and 15 is about 13 the rate between years 1 and 5.

The average rate of change between years 11 and 15 is about 3 times the rate between years 1 and 5.

The average rate of change between years 11 and 15 is about 12 the rate between years 1 and 5 .

The average rate of change between years 11 and 15 is about 2 times the rate between years 1 and 5 .

Answer 1

Given that the population is modeled by:
f(x)=125(0.9)^x
the average rate of change will be as follows:
Year 11 and 15:
f(11)=125(0.9)^11
=39.226
f(15)=125(0.9)^15
=25.736
rate of change=(25.736-39.226)/(15-11)
=-3.37
year 1 and 5
f(1)=125(0.9)^1
=112.5

f(5)=125(0.9)^5
=73.811
rate of change:
(73.811-112.5)/4
=-9.67

dividing the two results of rate of change we get:
-9.67/-3.37
=2.97-=3
From the above we conclude that the average rate of change between year 1 to 5 is above 3 times that of year 11 to year 15

Answer 2

The correct answer is:

The average rate of change between years 11 and 15 is about 1/3 the rate between years 1 and 5.

Step-by-step explanation:

The amount at year 1 is 125(0.9)¹ = 112.5. The amount at year 5 is 125(0.9)⁵ = 73.811. This means the average rate of change is

(73.811-112.5)/(5-1) = -38.689/4 = -9.67225.

The amount at year 11 is 125(0.9)¹¹ = 39.226. The amount at year 15 is 125(0.9)¹⁵ = 25.736. This means the average rate of change is

(25.736-39.226)/(15-11) = -13.49/4 = -3.3725. This is around 1/3 of the rate of change from years 1 to 5.